$x^3 + y^3 = 7 \implies x = \sqrt[3]{7 - y^3}.$

$And\ \dfrac{1}{x} + \dfrac{1}{y} = -\ \dfrac{1}{2} \implies \dfrac{2}{x} + \dfrac{2}{y} = -\ 1 = \dfrac{- y }{y}\implies \dfrac{2}{x} = -\ \dfrac{y + 2}{y} \implies x = -\ \dfrac{2y}{y + 2}.$

$So\ -\ \dfrac{2y}{y + 2} = \sqrt[3]{7 - y^3} \implies \dfrac{8y^3}{y^3 + 6y^2 + 12y + 8} = y^3 - 7.$

Now that implies a polynomial of degree 6, but the coefficient of y^{6} is 1 so the integer root theorem applies. The constant term is - 56.

So if y is an integer $y = \pm 56, \pm 28, \pm 14, \pm 8, \pm 7, \pm 4, \pm 2, or\ \pm 1.$

Furthermore, we are looking for an integer x so y^{3} - 7 must be a perfect cube. Some numerical exploration shows that y = 2 works as does

y = - 1.

$y = 2 \implies x^3 + 2^3 = 7 \implies x^3 = -\ 1 \implies x = -\ 1.\ And\ \dfrac{1}{-1} + \dfrac{1}{2} = -\ \dfrac{1}{2}.$

$y = - 1 \implies x^3 + (-1)^3 = 7 \implies x^3 = 8 \implies x = 2.\ And\ \dfrac{1}{2} + \dfrac{1}{-1} = -\ \dfrac{1}{2}.$

Solutions are: $(-1, 2)\ or\ (2, -1).$